Difference between revisions of "2005 AMC 12A Problems/Problem 15"
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<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math> | <math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math> | ||
− | [[Image:2005_12A_AMC-15.png]] | + | [[Image:2005_12A_AMC-15.png]] |
== Solution == | == Solution == | ||
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Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. | Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. | ||
− | Now we have to find <math>CF</math>. Notice <math>\triangle OCD \ | + | Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]: |
<div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div> | <div style="text-align:center;"><math>\frac{OF}{OC} = \frac{OC}{OD}</math><br /><math>\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}</math><br /><math>OF = \frac 19r</math></div> | ||
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{{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}} | {{AMC12 box|year=2005|num-b=14|num-a=16|ab=A}} | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 15:19, 23 September 2007
Problem
Let be a diameter of a circle and
be a point on
with
. Let
and
be points on the circle such that
and
is a second diameter. What is the ratio of the area of
to the area of
?
Solution
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or (
is the foot of the perpendicular from
to
).
Call the radius . Then
,
. Using the Pythagorean Theorem in
, we get
.
Now we have to find . Notice
, so we can write the proportion:
![$\frac{OF}{OC} = \frac{OC}{OD}$](http://latex.artofproblemsolving.com/0/2/2/02291691270e57eea057ea561416318b8de9fc01.png)
![$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$](http://latex.artofproblemsolving.com/6/e/1/6e1dcd67ba34e71f9f880510df1fd482941f6e37.png)
![$OF = \frac 19r$](http://latex.artofproblemsolving.com/5/a/4/5a4dcf36a069d28f1265cf8c1b825c9ae46c0cbd.png)
By the Pythagorean Theorem in , we have
.
Our answer is .
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |